Electrical Circuits

~Moving Charge Put to Use

The Circuit• All circuits, no matter how simple or complex, have one

thing in common, they form a complete loop.• As mentioned before, circuits should have various circuit

elements in the loop.

Series Circuit

• Have you ever driven down a 1 lane road? • You can keep moving until…• If there is an accident all traffic stops, there is no other

road to follow.

• A series circuit is similar to a one lane road, current can flow in only one path.

• Even if you add a 2nd resistor in series, there is still just 1 path.

Series Circuit

R1

R2

Series Circuit• One path means all components have the same

current• The Voltage provided by the source must equal the

Voltage drop across the resistor(s)

source resistorV V

RVI

Series Circuit• One path means all components have the

same current• What is the voltage drop across R1?

1 2source R RV V V

R1

V

R2

I 1 1RV IR

2 2RV IR

Series Circuit• How do we find Req?

1 2series R RV V V

R2

R1

V I

e 1 2Rs q s sI I R I R

e 1 2R q R R Divide both sides by Is

ReqV

The Series Circuit (cont.)• Every series configuration can be reduced to a single

value for resistance known as the equivalent resistance, or Req.

• The formula for Req is as follows for series:

• This can be used as a step to solve for the current in the circuit or the voltage across each resistor.

R1

R2

Req

1 2eqR R R

I

Sample Problem (Series)• A circuit is configured in series as shown below.

– What is the equivalent resistance (Req)?

– What is the current through the circuit?

(Hint: Use Ohm’s Law.)

10

20

30

6V

1 2 3eqR R R R 10 20 30eqR 60eqR

eq eqeq eq

eq eq

V VR I

I R

6

60eq

VI

0.1eqI A

Ieq = 0.1A

606V

Sample Problem (Series) (cont.)• We still have one question to ask. What are the voltages

across each resistor?

– For the 10 Resistor:

– For the 20 Resistor:

– For the 30 Resistor:

• What do you notice about the

voltage sum?

10

20

30

6V

Ieq = 0.1A

VR V IR

I

V IR 0.1 10V A

0.1 20V A

0.1 30V A

1V V

2V V

3V VV IR

V IR

Voltages across resistors in series add to make up the total voltage.

1 2 3 6V V V V 6 eqV V

Series Circuit Summary• Current is constant throughout the entire circuit.

• Resistances add to give Req.

• Voltages across each resistor add to give Veq.

1 2eqI I I

1 2eqR R R

1 2eqV V V

Devices that Make Use of the Series Configuration

• Although not practical in every application, the series connection is crucial as a part of most electrical apparatuses.– Switches

• Necessary to open and close entire circuits.

– Dials/Dimmers• A type of switch containing a variable resistor

(potentiometer).

– Breakers/Fuses• Special switches designed to shut off if current is too

high, thus preventing fires.

– Ammeters• Since current is constant in series, these current-

measuring devices must be connected in that configuration as well.

The Parallel Circuit (cont.)

• Parallel circuits are similar to rivers with branches in them.

• The current from the river divides into multiple paths.• After the paths, the water recombines into the same

amount of flowing water.1 2eqI I I

IeqIeqI1

I2

• A parallel circuit is similar to a river that branches, current can flow in multiple paths.

• Once the paths end, the total flow remains the same

Parallel Circuit

R2R1

The Parallel Circuit• Notice that the circuit branches out to each resistor,

allowing multiple paths for current to flow.

• If there are exactly two clear paths from the ends of one resistor to the ends of the other resistor.

R1 R2

Branch

X

BranchX

A break in one of the branches of a parallel circuit will not disable current flow in the remainder of the circuit.

Parallel Circuit• How do we find Req a parallel circuit?

1 2parallel R RI I I

R2

V

e 1 2Rp p p

q

V V V

R R

e 1 2

1 1 1

R q R R

Divide both sides by Vp

ReqVR1 R2

V

Use Ohm’s law

The Parallel Circuit (cont.)• Notice how every resistor has a direct connection to the

DC source. This allows the voltages to be equal across all resistors connected this way.

• An equivalent resistance (Req) can also be found for parallel configurations. It is as follows:

R1 R2

1 2eqV V V

Req

1 2

1 1 1

eqR R R

Sample Problem (Parallel)• A circuit is configured in parallel as shown below.

– What is the equivalent resistance of the circuit?

30 6V

1 2 3

1 1 1 1

eqR R R R

1 1 1 1

30 30 60eqR

11 1 130 30 60

eqR

12eqR

126V

Sample Problem (Parallel)• What is the current in the entire circuit?

• What is the current across each resistor?

30 6V

eq eqeq eq

eq eq

V VR I

I R 6

12eq

VI

0.5eqI A

VI

R

6

30

VI

6

60

VI

0.2I A 0.1I A

The 30 Resistors The 60 Resistor

Parallel Circuit Summary• There are several facts that you must always keep in

mind when solving parallel problems.– Voltage is constant throughout the entire parallel circuit.

– The Inverses of the Resistances add to give the inverse of Req.

– Current through each resistor adds to give Ieq.

– Make use of Ohm’s Law.V VR I V IR

I R

1 2eqV V V

1 2eqI I I

1 2

1 1 1

eqR R R

Devices that Make Use of the Parallel Configuration

• Although not practical or safe in every application, the parallel circuit finds definite use in some electrical apparatuses.– Electrical Outlets

• Constant voltage is a must for appliances.

– Light Strands• Prevents all bulbs from going out when a single

one burns out.

– Voltmeters• Since voltage is constant in parallel, these

meters must be connected in this way.

Lights demo• DC source with 3 lights in series

• DC source with 3 lights in parallel

• DC source with 2 lights in series 1 parallel

• DC source with 1 lights in series 2 parallel

Combination Circuits• Some circuits have series/parallel combinations • These can be reduced using equivalent resistance

formulas.• Now let’s solve a problem involving this circuit.

R1R2

R3 R4

SeriesParallel

Sample Problem (Combo)

What is the equivalent resistance (Req) of the circuit?– First, we must identify the various combinations present.

Series

Parallel

1 2eqR R R 10 30eqR

40eqR 1 2

1 1 1

eqR R R

1 1 1

20 20eqR

10eqR

Series Parallel

1040

301020 20

25V

Sample Problem (Combo)• The simplified circuit only shows the equivalent

resistances. Is the circuit now fully simplified?• Now, we must identify the final configuration.

Series

Parallel

1040

301020 20

25V

4010

25V1 2eqR R R 40 10eqR 50eqR

Series

50

Sample Problem (Combo)• The circuit is further simplified below. Can it be

simplified again?• Now, the circuit is completely simplified.• What is the current in the entire circuit?

4010

25V

Series

50

5025V

eq eqeq eq

eq eq

V VR I

I R 25

50eq

VI

0.5eqI A

Combination Circuits

• Parallel Paths: Must make a complete loop through two resistors with out touching any other component.

• Series Paths: Must form a path through multiple resistors with out crossing an intersection.

Kichoff's current laws

1

• Choose a direction and label the current in each branch• Identity the number of unknowns as develop as many equation

• Label the polarity of each Vr for all resistors.

• Apply Kirchoff’s junction rule (sum of current in equals sum of current out.• Apply Kirchoff’s loop rule. The sum of all voltages around a loop must equal zero• Solve the simultaneous equations

40V

30 1 35V

10

10

I1

I2

I3

Kichoff's current laws

140V

30 1 35V

10

10

25V3010

30

15V

• Make a terminal voltage slide

25V3010

30

15V

24μf

12μf

36μf

220V

114

117

220V

114

117

140